On Mon, 2005-04-18 at 15:35 +1000, Bernard Pope wrote:
> On Mon, 2005-04-18 at 01:08 -0400, David Menendez wrote:
> > Trevion writes:
> >
> > > On 4/18/05, Lloyd Allison <[EMAIL PROTECTED]>
> > wrote:
> > > > Is it possible to define Y in Haskell (pref' H98) --
> > > > and get it to pass type chec
On Mon, 2005-04-18 at 01:08 -0400, David Menendez wrote:
> Trevion writes:
>
> > On 4/18/05, Lloyd Allison <[EMAIL PROTECTED]>
> wrote:
> > > Is it possible to define Y in Haskell (pref' H98) --
> > > and get it to pass type checking?
> >
> > I believe that
> >
> > y f = f (y f)
> >
> > is the
Trevion writes:
> On 4/18/05, Lloyd Allison <[EMAIL PROTECTED]>
wrote:
> > Is it possible to define Y in Haskell (pref' H98) --
> > and get it to pass type checking?
>
> I believe that
>
> y f = f (y f)
>
> is the normal way to do it.
I've also seen this:
y f = g
where g = f g
--
On 4/18/05, Lloyd Allison <[EMAIL PROTECTED]> wrote:
> Is it possible to define Y in Haskell (pref' H98) --
> and get it to pass type checking?
I believe that
y f = f (y f)
is the normal way to do it.
.trev
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Is it possible to define Y in Haskell (pref' H98) --
and get it to pass type checking?
--
E.g. As in SML
let
datatype 'a rt = recrt of ('a rt) -> 'a;
fun Y G= let fun Ggg (recrt g) n = G(g (recrt g))n
in Ggg (recrt Ggg)
end;
fun F f n = if n=0 then 1 else n*f(n-1
